In this screencast we are going to learn how to approach a material balance. The most important
thing to realize is, when balancing steady state systems, that mass in always equals
mass out. It does not matter whether there is a reaction. It does not matter if there
is more than one phase. Doesn't matter the mole fractions of anything. Mass in has to
equal mass out. Here are some pointers. That might be able to help you. The major problem
is that students don't know how to take a problem and turn it into a system. So ask
yourself what does the problem ask for? Write it in terms of variables that you set. Read
the problem carefully. Ask yourself “What do they want?". Look for key words or phrases.
If you see something like overhead product. Then it means that you have product coming
out of the top of perhaps your column. Be careful of what you assume. Some examples
of this are adding streams, leaving streams out, and let me show you a typical example.
Let’s say we have an absorber. There is a gas stream that comes in, comes out, and
a pure liquid enters the absorber to absorb some of the components of the gas stream.
Well if this stream enters the system. It needs to exit the system. Even if the problem
doesn't specifically say that the there is an aqueous stream exiting the system. It has
to come out even if it is not pure. In fact it probably isn't, because the purpose is
for it to pick up some of the components from the gas stream. The next thing to do is draw
the picture. Label the unknowns and the knowns. What I tend to do is use “m” for mass,
and I use “n” for moles. For simpler problems I will use x for mass fractions, and y for
our mole fractions. If I am looking for particular mass fraction or mole fraction I tend to label
with something that indicates the components. For example if I have a mass fraction of benzene
I will call this x sub B (x_B). Finally if there is more than one stream that has benzene.
Let’s say there is benzene in stream 2 and in stream 4, I will write x sub B, 2 (x_B2).
The more organized you are, the easier it is for you to solve. Another suggestion is
to solve for the easy unknowns ahead of time. Don't do major calculations because, first
of all, you might end up confusing yourself, and second of all, somebody looking at your flow
chart may not understand how you got a certain number, but things like if you have a two
component system, and it says that it has a mole fraction of 40% benzene, and the rest
toluene. You can write 0.40 Benzene, and 0.60 Toluene. Now, we state with the material balance,
and the thing to remember is that there is one mass balance for every component. One
thing that you can do is that you can substitute one overall balance for a component balance,
and one of the things that you have to be careful of when you are doing your mass balances,
is that each one of your equations or balances is independent. So let me show you an example.
We have 2 streams that enter a column. Our stream 1 is 30 % A, and the rest B. There
is an example of solving for simple unknowns. Stream 2 is 40% A and the rest C. So two streams
exit the column. For stream 3 it is 100% A, and stream 4 is 20% B and the rest A and C.
We are going to write this using the fact that the sum of the mole fractions equals 1.
So our xc_4 is equal to 0.80-xa_4, and we are going to assume that everything is
in mass fractions. So we need to write the mass material balances. We will call this
“m1”, “m2”, “m3”, and “m4”. So now lets do these mass balances, and we
will start with A. So the amount of A that comes in is 30% of “m1” plus 40% of “m2”,
and exiting we have 100% of “m3” or just “m3” plus our xA_4 times “m4”. Now
we will do the B balance. Coming in the first stream 70% is “m1”. There is nothing in
“m2”. There is nothing in “m3”, it is 20% of “m4”. Now our C balance, which
nothing is in “m1”, C is 60% of “m2”. Coming out we have (0.8-xa_4)*m4. These are
our three component balances. Let’s see what happens if we add them all together.
So note that when we add them all together, what we get is the overall balance. So we
can only use three out of these 4 equations to solve our system. Finally note that we
have 1, 2, 3, 4, 5 unknowns, but we only have 3 mass balances that we can write. So as the
system is stated we cannot solve it. This is what is called a degree of freedom analysis,
and I would like to direct you to another screencast that describes what a degree of
freedom analysis is.